Solve for x, y
x=1200
y=1400
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5x-3y=1800,6x-4y=1600
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
5x-3y=1800
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=3y+1800
Add 3y to both sides of the equation.
x=\frac{1}{5}\left(3y+1800\right)
Divide both sides by 5.
x=\frac{3}{5}y+360
Multiply \frac{1}{5} times 1800+3y.
6\left(\frac{3}{5}y+360\right)-4y=1600
Substitute \frac{3y}{5}+360 for x in the other equation, 6x-4y=1600.
\frac{18}{5}y+2160-4y=1600
Multiply 6 times \frac{3y}{5}+360.
-\frac{2}{5}y+2160=1600
Add \frac{18y}{5} to -4y.
-\frac{2}{5}y=-560
Subtract 2160 from both sides of the equation.
y=1400
Divide both sides of the equation by -\frac{2}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{3}{5}\times 1400+360
Substitute 1400 for y in x=\frac{3}{5}y+360. Because the resulting equation contains only one variable, you can solve for x directly.
x=840+360
Multiply \frac{3}{5} times 1400.
x=1200
Add 360 to 840.
x=1200,y=1400
The system is now solved.
5x-3y=1800,6x-4y=1600
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1800\\1600\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right))\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right))\left(\begin{matrix}1800\\1600\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-3\\6&-4\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right))\left(\begin{matrix}1800\\1600\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\6&-4\end{matrix}\right))\left(\begin{matrix}1800\\1600\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-\left(-3\times 6\right)}&-\frac{-3}{5\left(-4\right)-\left(-3\times 6\right)}\\-\frac{6}{5\left(-4\right)-\left(-3\times 6\right)}&\frac{5}{5\left(-4\right)-\left(-3\times 6\right)}\end{matrix}\right)\left(\begin{matrix}1800\\1600\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2&-\frac{3}{2}\\3&-\frac{5}{2}\end{matrix}\right)\left(\begin{matrix}1800\\1600\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 1800-\frac{3}{2}\times 1600\\3\times 1800-\frac{5}{2}\times 1600\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1200\\1400\end{matrix}\right)
Do the arithmetic.
x=1200,y=1400
Extract the matrix elements x and y.
5x-3y=1800,6x-4y=1600
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
6\times 5x+6\left(-3\right)y=6\times 1800,5\times 6x+5\left(-4\right)y=5\times 1600
To make 5x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 5.
30x-18y=10800,30x-20y=8000
Simplify.
30x-30x-18y+20y=10800-8000
Subtract 30x-20y=8000 from 30x-18y=10800 by subtracting like terms on each side of the equal sign.
-18y+20y=10800-8000
Add 30x to -30x. Terms 30x and -30x cancel out, leaving an equation with only one variable that can be solved.
2y=10800-8000
Add -18y to 20y.
2y=2800
Add 10800 to -8000.
y=1400
Divide both sides by 2.
6x-4\times 1400=1600
Substitute 1400 for y in 6x-4y=1600. Because the resulting equation contains only one variable, you can solve for x directly.
6x-5600=1600
Multiply -4 times 1400.
6x=7200
Add 5600 to both sides of the equation.
x=1200
Divide both sides by 6.
x=1200,y=1400
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}